Question

Find the number of ways in which \[35\] apples can be distributed among 3 boys so that each can have any number of apples.
A. \[1332\]
B. \[666\]
C. \[333\]
D. None of these

Answer 1

Hint: To calculate the possible number of ways, use the formula of the number of ways in which \[n\] objects can be distributed among \[r\] people that each can get any number of objects. Then, solve it by applying the combination formula and get the required answer.



Formula Used:The number of ways in which \[n\] objects can be distributed among \[r\] people that each can get any number of objects: \[{}^{n + r - 1}{C_{r - 1}}\]
The combination formula: \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Factorial: \[n! = 1 \times 2 \times ... \times n = n\left( {n - 1} \right)!\]



Complete step by step solution:Given:
Total number of apples: \[35\]
Number of boys: 3

We have to calculate the number of ways in which \[35\] apples can be distributed among 3 boys so that each can have any number of apples.
Now apply the formula the number of ways in which \[n\] objects can be distributed among \[r\] people so that each can get any number of objects \[{}^{n + r - 1}{C_{r - 1}}\].
So, the possible number of ways are:
\[{}^{35 + 3 - 1}{C_{3 - 1}} = {}^{37}{C_2}\]
Solve it by applying the combination formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\].
\[{}^{35 + 3 - 1}{C_{3 - 1}} = \dfrac{{37!}}{{2!\left( {37 - 2} \right)!}}\]
\[ \Rightarrow {}^{35 + 3 - 1}{C_{3 - 1}} = \dfrac{{37 \times 36 \times 35!}}{{2!35!}}\]
\[ \Rightarrow {}^{35 + 3 - 1}{C_{3 - 1}} = \dfrac{{37 \times 36}}{2}\]
\[ \Rightarrow {}^{35 + 3 - 1}{C_{3 - 1}} = 666\]
Thus, the number of ways in which \[35\] apples can be distributed among 3 boys so that each can have any number of apples is \[666\].



Option ‘B’ is correct

Note: In this type of question, sometimes students directly apply the combination formula. Because of that, they get the wrong answer.